Verbal subgroup equals power subgroup in abelian group

This article gives a proof/explanation of the equivalence of multiple definitions for the term verbal subgroup of abelian group
View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose $G$ is an abelian group and $H$ is a verbal subgroup. Then, there exists an integer $n$ such that $H$ is precisely the set of $n^{th}$ multiples in $G$:

$H=\{nx\mid x\in G\}$.

Conversely, the set of $n^{th}$ multiples form a verbal subgroup for any integer $n$.

Related facts

• Verbal upper-hook verbal equals verbal in abelian group
• Intersection of two verbal subgroups of abelian group is verbal in both and in whole group

Proof

Verbal subgroup implies it is the set of nth powers/multiples

Given: A group $G$, a verbal subgroup $H$ with a collection $W$ of words generating it.

To prove: There exists a natural number $n$ such that $H$ is the set of $n^{th}$ multiples.

Proof:

1. Let $w\in W$. We first prove that there exists an integer $n_w$ such that an element of $G$ can be expressed using the word $w$ if and only if it is a $n_w^{th}$ multiple: Note first that by Abelianness, we can replace $w$ by a word (written additively) of the form $\sum _{i=1}^r{a}_i{x}_i$, where $a_i$ are integers and $x_i$ are indeterminates. Let $n_w$ be the gcd of all the $a_i$s. We claim this $n_w$ works.
   • There exist integers $m_i$ such that $\sum _{i=1}^r{a}_i{m}_i=n_w$. Thus, if $y=n_{w}x$, we get $y=\sum _{i=1}^r{a}_i(m_{i}x)$, so $y$ can be written using the word $w$.
   • Conversely, if $y$ can be written using the word $w$, then $y=\sum _{i=1}^r{a}_i{x}_i=\sum _{i=1}^r{n}_w(a_i/n_w)x_i=n_w\sum _{i=1}^r(a_i/n_w)x_i$, which is clearly a multiple of $n_w$.
2. We now claim that the gcd of all the $n_w,w\in W$, is precisely the $n$ that we seek:
   • Note that each element expressible using a word $w\in W$ is a multiple of $n_w$, and thus, a multiple of $n$. Thus, any element in the subgroup generated by such elements is also a multiple of $n$.
   • Conversely, since $n$ is the gcd of the $n_w$, there exist $w_1,w_2,\dots ,w_s$ and integers $b_1,b_2,\dots ,b_s$ such that $\sum b_i{n}_{w_i}=n$. Thus, if $y=nx$, $y=\sum b_i(n_{w_i}x)$, hence $y$ is in the subgroup generated by multiples of $n_w$.

Set of nth powers is a verbal subgroup

Since $G$ is Abelian, the subset is a subgroup. Moreover, since $nx=x+x+\dots +x$ is itself a word, we see that it is a verbal subgroup.